A current of 16A is divides between 2 branches in parallel Of resistance 8ohm and 12 ohm respectively. The currents are ​

Réponse:

current in each branch is 6.4 A and 9.6 A.

Explanation:

Given that,

Current I = 16 A

First resistance R_{1}=8\text{\O}megaR

1

=8Ømega

Second resistance R_{2}=12\text{\O}megaR

2

=12Ømega

We know that,

The Current in 8 ohm,

I_{1} = \dfrac{R_{1}}{R_{1}+R_{2}}\times I_{total}I

1

=

R

1

+R

2

R

1

×I

total

I_{1} = \dfrac{8}{20}\times16I

1

=

20

8

×16

I_{1}= 6.4\ AI

1

=6.4 A

Now, The current in 12 ohm

I_{2}=16-6.4I

2

=16−6.4

I_{2}= 9.6\ AI

2

=9.6 A

Hence, The current in each branch is 6.4 A and 9.6 A.